I ran across an interesting math mistake in a post by Bob Murphy on the blog Crash Landing. He quotes from Peter L. Bernstein’s 1993 book Capital Ideas (which is apparently “A savvy appreciation of how a small band of disinterested academics has revolutionized the way Wall Street and its offshore counterparts manage the world’s investment wealth,” according to Kirkus Review).

Anyway, here’s the quote:

[Cowles] must have been a fiendish bridge player. Here is one passage from his notes on the game:

If each of 50 million bridge players in the US plays 200 sessions of 40 deals each, this adds up to 50 million*200*40 = 400 billion hands dealt each in US (sic). The probabilities on any given hand being dealt with 13 cards of one suit are .00000000000156. The chances of a hand with 13 cards of one suit being dealt in the US in any given year, therefore are 400 billion times .00000000000156=.624.

The challenge posed was to find the math mistake. I’m pretty sure I know which one he was referring to, but I’ll share my train of thought anyway.

My first guess was that it has to do with the 400 billion deals going on in the US each year. But in retrospect, I think this number is correct given the assumptions [and assuming that it’s only the dealer’s hand that’s being looked at; there are four people playing with each deal]. What I disagree with is that those assumptions are reasonable. Are there really 50 million bridge players in the US? Maybe. But the US Census estimates that there are only 301,621,157 people in the US as of July 1, 2007; that means that 1 in 6 people is a bridge player. It might be true that 1/6 of the folk in the US know how to play bridge (maybe), but I’m pretty sure that it is not the case that 1/6 of the entire US population deals 8000 hands per year. That’s almost 22 hands per day (and that only counts when that person is the dealer!), each and every day, for an entire year. If that were the case, who would have time for blogging?

So that 400 billion is way off, but because of unreasonable assumptions rather than an actual mistake (and not having read the actual book, it’s possible he knew that he was wildly overestimating the number of hands dealt). Then I turned my attention to the line “The probabilities on any given hand being dealt with 13 cards of one suit are .00000000000156.” This number comes from the fact that the number of ways to choose 13 cards from a deck of 52 is (52 choose 13), or approximately 6.35×10^{11}. There’s only one way to get all 13 cards in a particular suit (say, spades) so the chances of getting that or any particular configuration of cards is 1/(52 choose 13) or approximately 1.57×10^{-12}. So that number fits the paragraph above, if you interpret “a suit” to mean a particular suit. If you just want all the cards to be of the same suit, then you’d have to multiply that probability by 4.

Which means that probability is probably wrong too, but it does depend on interpretation. Finally I turned my attention to last line, about multiplying the probability by 400 billion to find the likelihood of getting all cards in the same suit because there are 400 billion hands dealt. And this, I am certain, is the error that Bob Murphy was referring to. As Bob’s brother pointed out, if they had used a population that was large enough (or indeed, simply multiplied the tiny probability above by 4 to take into account that there are 4 different suits in the deck), the probability of getting all 13 cards in the same suit would appear to be more than one.

The correct way to solve this would be to find the likelihood that a particular hand dealt was not all of a given suit (1-1.57×10^{-12}), raise that to the 400 billion power to find the probability that none of those 400 billion hands dealt were all of a suit, and then subtracting that from 1. Using Excel, it appears that there’s a 99.996% change that someone, somewhere got that particular configuration of cards.

What a rich little paragraph! It looks like Bernstein has written a sequel to the book in 2005; let’s hope that in this version, the math was checked a bit more carefully.

December 2, 2008 at 8:52 pm |

For

smallprobabilities, though, that approximation is pretty close to being right. That is, 1-(1-p)n is approximately np when np is much less than 1. Certainly it introduces a lot less error than the other assumptions.December 2, 2008 at 8:53 pm |

In the previous comment, I meant to write 1-(1-p)^n. (Apparently only certain HTML tags are allowed in comments.)

December 3, 2008 at 10:38 am |

Good point Michael!